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3 years ago

8

On a hot summer day, you need to leave your house for 30 minutes. To use the least amount of energy, should you turn off your ai

r conditioning system, turn up the thermostat, or leave it set at your normal temperature?

1 answer:

Kobotan [32]3 years ago

5 0

Leave it at your normal temperate, if you turn it off it'll take more energy for it to cool off when u get back home.

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Nina and Jon are practicing an ice skating routine. Nina is standing still. Jon, who is twice as heavy as Nina, skates toward he

Harman [31]

Answer:

A

Explanation:

• Nina experiences a force equal to f.

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3 years ago

What is the role of equations in this course?

Karolina [17]

That will depend on which course you're talking about. It will be a minor role in, say, Maritime Law or Comparitive Religion, but a major one in, say, Particle Physics or Linear Algebra.

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3 years ago

A carbon rod with a radius of 1.9 mm is used to make a resistor. What length of the carbon rod should be used to make a 3.7 Ω re

vladimir2022 [97]

Answer:

Length = 2.32 m

Explanation:

Let the length required be 'L'.

Given:

Resistance of the resistor (R) = 3.7 Ω

Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]

Resistivity of the material of rod (ρ) = $1.8\times 10^{-5}\ \Omega\cdot m$

First, let us find the area of the circular rod.

Area is given as:

$A=\pi r^2=3.14\times (0.0019)^2=1.13\times 10^{-5}\ m^2$

Now, the resistance of the material is given by the formula:

$R=\rho( \frac{L}{A})$

Express this in terms of 'L'. This gives,

$\rho\times L=R\times A\\\\L=\frac{R\times A}{\rho}$

Now, plug in the given values and solve for length 'L'. This gives,

$L=\frac{3.7\ \Omega\times 1.13\times 10^{-5}\ m^2}{1.8\times 10^{-5}\ \Omega\cdot m}\\\\L=\frac{4.181}{1.8}=2.32\ m$

Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.

5 0

3 years ago

Who wins a tug-of-war one who pushes harder at the ground or pulls harder at the rope

mixas84 [53]

I think pulls harder on the rope

4 0

3 years ago

Read 2 more answers

9. How much work is done when a 15kg box is lifted to a height of 2 meters?

torisob [31]

Answer: W = 294 J

Explanation: Solution:

Work is expressed as the product of force and the distance of the object.

W = Fd where F = mg

W= Fd

= mg d

= 15 kg ( 9.8 m/s²) ( 2m )

= 294 J

5 0

3 years ago